$\binom{3n}{n}$ divisible by $2016$ If $\dbinom{3n}{n}$ divisible by $2016$ then, what is smallest value of positive number of $n$?
Notes: I can find $n=23$. $2016=2^5\cdot 3^2 \cdot 7$ We can give to $n=1,2,\dots ,22$ and $2016 \not| \dbinom{3n}{n}$. For $n=23$, $\dbinom{69}{23}$ divisible by $2^5, 3^2$ and $7$.
I care to shortcut.
 A: There is a small shortcut available as the highest possible multiplicity of any prime $p$ in $\binom mn$ is the largest power of of $p$ no greater than $m$. It's further limited to the largest power of $p$ in the falling factorial part of the expression, the values from $m{-}n{+}1$ to $m$. In this case you can skip the values where there is no sufficiently high power of two in the interval $[2n{+}1, 3n]$, so you could check $[11..15]$, then $22$ onwards. 
For the actual calculation, the final values of the binomial coefficients are not required. To find $v_p(n!)$, the multiplicity of $p$ in $n!$,  calculate $\displaystyle v_p(n!)=\sum_{i=1}^k\left\lfloor \frac n{p^i}\right\rfloor$ (where $k$ is big enough to include all powers of $p$ up to $n$). Then you can check the appropriate powers for $2016=2^5\cdot 3^2\cdot 7^1$, calculating for example $v_2(\binom {3n}{n})= v_2((3n)!)-v_2((2n)!)-v_2(n!)$
The following values of $n<100$ have $\displaystyle 2016 \mid \binom {3n}n$ : $23, 31, 43,$ $46, 47, 55,$ $59, 62, 86,$ $87, 91, 92,$ $93, 94, 95$
